Which of the following statements is/are true? (A) (tan^-1y-x)dy = (1…

The correct answer is Option (2) → (B) and (C) only

An equation \(\frac{dy}{dx} = F(x, y)\) is homogeneous when \(F(\lambda x, \lambda y) = F(x, y)\) for every \(\lambda\); it is linear (first order) when it can be written as \(\frac{dy}{dx} + Py = Q\) with P, Q functions of x only.

(A) \((\tan^{-1}y-x)dy = (1 + y^2)dx\) is a differential equation where variables are separable. — Incorrect. Solving for \(\frac{dx}{dy}\):

    \(\displaystyle \frac{dx}{dy} = \frac{\tan^{-1}y - x}{1 + y^2}\)

The right side is a difference involving x and y, not a product g(x)h(y), so the variables cannot be separated. It is a linear equation in x:

    \(\displaystyle \frac{dx}{dy} + \frac{1}{1 + y^2}\,x = \frac{\tan^{-1}y}{1 + y^2}\)

(B) \((1+x^2)dy+ 2xy\,dx = \cot x\,dx \;(x \ne 0)\) is a first order linear differential equation. — Correct. Dividing by \((1 + x^2)\,dx\):

    \(\displaystyle \frac{dy}{dx} + \frac{2x}{1 + x^2}\,y = \frac{\cot x}{1 + x^2}\)

This is of the form \(\frac{dy}{dx} + Py = Q\), with P and Q functions of x only.

(C) \((4x+6y+5)dy - (3y+2x+4)dx = 0\) is not a homogeneous differential equation. — Correct. Writing it as \(\frac{dy}{dx} = F(x, y)\):

    \(\displaystyle \frac{dy}{dx} = \frac{2x + 3y + 4}{4x + 6y + 5} = F(x, y)\)

    \(\displaystyle F(\lambda x, \lambda y) = \frac{2\lambda x + 3\lambda y + 4}{4\lambda x + 6\lambda y + 5}\)

Because of the constants 4 and 5, \(\lambda\) does not cancel, so \(F(\lambda x, \lambda y) \ne F(x, y)\) and the equation is not homogeneous.

(D) \((xy)dx-(x + y^2)dy = 0\) is a homogeneous differential equation. — Incorrect. Writing it as \(\frac{dy}{dx} = F(x, y)\):

    \(\displaystyle \frac{dy}{dx} = \frac{xy}{x + y^2} = F(x, y)\)

    \(\displaystyle F(\lambda x, \lambda y) = \frac{\lambda^2 xy}{\lambda x + \lambda^2 y^2} = \frac{\lambda xy}{x + \lambda y^2}\)

This is not equal to F(x, y) (the terms x and y² have different degrees), so the equation is not homogeneous.

So only (B) and (C) are true.